- #1
Tomer
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Thanks for reading!
It's actually a problem in Physics, which is analogical to a problem in probability:
Given are two pieces of paper with area A, one black and one white. we then cut the black one to n pieces, so that each piece has the area A/n (equal sized)
We then let the "little" (n will soon be going to infinity) black pieces fall from above on the other white piece of paper, with the limitation that they have to cover the paper only, and that they either cover "new white spots" on the paper, or align with previously scattered pieces.
I ought to find the expectation value of the surface *not covered* by n pieces.
I then ought to find the limit of this, when n goes to infinity.
[itex]Lim_{n\rightarrow\infty} (1+\frac{1}{n})^n = e[/itex]
(I've been given that as a hint).
I defined S[itex]_{w}[/itex] and S[itex]_{b}[/itex] to be random variables, describing the white and black surfaces remaining after scattering n black pieces.
So:
[itex]P(S_{w} = k\frac{A}{n}) = P(S_{b} = (n-k)\frac{A}{n}) = 1(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{n-k-1}{n})(\frac{n-k}{n})^k[/itex]
In the last step I just logically thought of the probability - if I want to cover k-n spots with black pieces using n black pieces, the first piece has probability 1 to cover any spot, the second has probability (1-1/n) to cover any other "free spot", and so on... until the (n-k)'th piece has the probability [itex]1-\frac{n-k-1}{n}[/itex] to cover a new white spot, since n-k-1 pieces are laying there, covering n-k-1 different spots. After "mannaging" to cover n-k black spots, I have to ensure that the remaining k pieces align with previous pieces. So each piece I throw has the probability [itex]\frac{n-k}{n}[/itex] to align with the n-k pieces now on the white paper. k times means [itex](\frac{n-k}{n})^k[/itex]
Assuming this is correct, all I need now is the limit of the expectation value of [itex]S_{w}[/itex], as n goes to infinity.
[itex]Lim_{n\rightarrow\infty} \sum^{n}_{k=1}(k\frac{A}{n})(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{n-k-1}{n})(\frac{n-k}{n})^k[/itex]
And this is where I'm stuck. I just don't know how to evaluate this. I'd appreciate tips. Of course, if something in my derivation is wrong I'd appreciate your help in correcting it.
Thanks A-LOT!
Tomer.
Homework Statement
It's actually a problem in Physics, which is analogical to a problem in probability:
Given are two pieces of paper with area A, one black and one white. we then cut the black one to n pieces, so that each piece has the area A/n (equal sized)
We then let the "little" (n will soon be going to infinity) black pieces fall from above on the other white piece of paper, with the limitation that they have to cover the paper only, and that they either cover "new white spots" on the paper, or align with previously scattered pieces.
I ought to find the expectation value of the surface *not covered* by n pieces.
I then ought to find the limit of this, when n goes to infinity.
Homework Equations
[itex]Lim_{n\rightarrow\infty} (1+\frac{1}{n})^n = e[/itex]
(I've been given that as a hint).
The Attempt at a Solution
I defined S[itex]_{w}[/itex] and S[itex]_{b}[/itex] to be random variables, describing the white and black surfaces remaining after scattering n black pieces.
So:
[itex]P(S_{w} = k\frac{A}{n}) = P(S_{b} = (n-k)\frac{A}{n}) = 1(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{n-k-1}{n})(\frac{n-k}{n})^k[/itex]
In the last step I just logically thought of the probability - if I want to cover k-n spots with black pieces using n black pieces, the first piece has probability 1 to cover any spot, the second has probability (1-1/n) to cover any other "free spot", and so on... until the (n-k)'th piece has the probability [itex]1-\frac{n-k-1}{n}[/itex] to cover a new white spot, since n-k-1 pieces are laying there, covering n-k-1 different spots. After "mannaging" to cover n-k black spots, I have to ensure that the remaining k pieces align with previous pieces. So each piece I throw has the probability [itex]\frac{n-k}{n}[/itex] to align with the n-k pieces now on the white paper. k times means [itex](\frac{n-k}{n})^k[/itex]
Assuming this is correct, all I need now is the limit of the expectation value of [itex]S_{w}[/itex], as n goes to infinity.
[itex]Lim_{n\rightarrow\infty} \sum^{n}_{k=1}(k\frac{A}{n})(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{n-k-1}{n})(\frac{n-k}{n})^k[/itex]
And this is where I'm stuck. I just don't know how to evaluate this. I'd appreciate tips. Of course, if something in my derivation is wrong I'd appreciate your help in correcting it.
Thanks A-LOT!
Tomer.